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center ((x+3)^2}{25}-\frac{(y-4)^2)/9 =1

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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simplify solve for inverse tangent line

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slope of tangent x², (3, 9)

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The slope of the tangent at a point is the derivative of the function at this point

ddx (x²)=2x

Find the value of 2x for x=3: 6

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

center ((x+3)^2}{25}-\frac{(y-4)^2)/9 =1

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